POJ-1840-Eqs【哈希】

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本文介绍了一种计算特定形式的五元立方方程解的数量的方法。通过预先计算两个未知数的组合，并使用哈希表存储结果，再遍历剩余三个未知数的所有可能组合，最终确定满足方程的解的数量。

Eqs
Description
Consider equations having the following form:
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.
Determine how many solutions satisfy the given equation.
Input
The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.
Output
The output will contain on the first line the number of the solutions for the given equation.
Sample Input
37 29 41 43 47
Sample Output
654

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
short hashs[25000001];
int main()
{
    int a1,a2,a3,a4,a5;
    int sum;
    while(cin>>a1>>a2>>a3>>a4>>a5)
    {
        memset(hashs,0,sizeof(hashs));
        for(int i=-50; i<=50; i++)
        {
            if(i==0)continue;
            for(int j=-50; j<=50; j++)
            {
                if(j==0)continue;
                sum=a1*i*i*i+a2*j*j*j;
                if(sum<0)
                    sum+=25000000;

                hashs[sum]++;
            }
        }
        int ans=0;
        for(int i=-50; i<=50; i++)
        {
            if(i==0)continue;
            for(int j=-50; j<=50; j++)
            {
                if(j==0)continue;
                for(int k=-50; k<=50; k++)
                {
                    if(k==0) continue;
                    sum=a3*i*i*i+a4*j*j*j+a5*k*k*k;
                    if(sum<0) sum+=25000000;
                    if(hashs[sum])
                        ans+=hashs[sum];
                }
            }
        }
        printf("%d\n",ans);
    }

}